php contradictions with $_POST

Question

I have this program:

if (!isset($_POST['foo'])) doSomeThing1();
else {
    if (!array_key_exists('foo',$_POST)) doSomeThing2();
    else doSomeThing3();
}

but... the program flow goes to the 3d case, failing with the error: undefined index 'foo' (in file.php, line xxx).

Could you explain, why? Why array_key_exists returns true (which brings the script to the 3d case) but subsequently it is "undefined index"?

have you tried looking at your data with a `var_dump($_POST)`? that could help us debuggin, too. also - what's in file.php:xxx? the code throwing the error could be relevant to debugging the error, don't you think? — Franz Gleichmann, Oct 17 '16 at 09:52
Please let us know what you are expecting when there is 'foo' key in $_POST; When there is NO 'foo' key in $_POST etc — Naresh Kumar Nakka, Oct 17 '16 at 09:55
Please `print` your post value first, even you can use `var_dump()` to undestand more that what data is coming exactly and why it is not meeting any condition. — Aman Garg, Oct 17 '16 at 09:54

score 1 · Answer 1 · edited Oct 17 '16 at 15:44

if (!isset($_POST['foo'])) doSomeThing1();
else {
    if (!array_key_exists('foo',$_POST)) doSomeThing2();
    else doSomeThing3();
}

As per this code how it's working is...

!isset($_POST['foo']) ==> returns true and executes dosomeThing1() when there is NO 'foo' key in $_POST array

if $_POST doesn't have any key it is checking !array_key_exists('foo',$_POST)

array_key_exists('foo', $_POST) checking whether 'foo' key is there in $_POST array or not. array_key_exists('foo', $_POST) is same as isset($_POST['foo']) so it is always executing doSomeThing3() when there is no 'foo' key in $_POST array.

Hope this explanation helps.

php contradictions with $_POST

1 Answers1